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anemone
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Let $ABCD$ be a convex quadrilateral such that $AB=BC,\,AC=BD,\,\angle CBD=20^{\circ},\,\angle ABD=80^{\circ}$. Find $\angle BCD.$
anemone said:Let $ABCD$ be a convex quadrilateral such that $AB=BC,\,AC=BD,\,\angle CBD=20^{\circ},\,\angle ABD=80^{\circ}$. Find $\angle BCD.$
anemone said:Hint:
Draw a line parallel to $BD$ from $C$ and see where that leads you...
anemone said:Let $ABCD$ be a convex quadrilateral such that $AB=BC,\,AC=BD,\,\angle CBD=20^{\circ},\,\angle ABD=80^{\circ}$. Find $\angle BCD.$
Albert said:$130^o$
MarkFL said:Please provide how you obtained your solution. :)
Albert said:step 1 :construct $\triangle ABC$ ,it is fixed , for $AB=BC$ , and $\angle ABC=20^o+80^o=100^o$
$\therefore \angle A=\angle C=40^o$
step 2: find point $D'$ on $AC$, and $\angle CBD'=20^o$
(step 1 and 2 can be done easily)
now point D must be on $BD'$ ray
step 3: construct circle $B$ with center $B$ and $BD=AC$
here point $D$ is the intersection of circle $B$ and $BD'$ ray
so $\angle BCD=\angle BCA+\angle ACD=40+90=130^o$
anemone said:Hi Albert!
When you mentioned of constructing a circle that centered at $B$, I am wondering if $r=BA\stackrel{\text{or}}{=}BD$?
Albert said:$r=BD=AC$
anemone said:Okay, thanks for your reply...but Albert, I have constructed the diagram and drew everything up according to suggestion, but, I failed to see why $\angle ACD=90^\circ$.
Albert said:$\angle ACD=90^o$
A convex quadrilateral is a four-sided polygon with all interior angles less than 180 degrees. This means that all of its vertices "point outwards" and the diagonal line segments connecting its vertices lie entirely within the shape.
To find $\angle BCD$ in a convex quadrilateral, you can use the formula: $\angle BCD = 180 - (\angle ABC + \angle ADC)$, where $\angle ABC$ and $\angle ADC$ are the adjacent angles to $\angle BCD$.
The sum of the interior angles in a convex quadrilateral is always 360 degrees. This is because a convex quadrilateral can be divided into two triangles, each with interior angles summing to 180 degrees, so the total sum is 360 degrees.
No, the measure of $\angle BCD$ cannot be greater than 180 degrees in a convex quadrilateral. As mentioned earlier, a convex quadrilateral has all interior angles less than 180 degrees, so $\angle BCD$ cannot exceed this limit.
Yes, there are two special cases for finding $\angle BCD$ in a convex quadrilateral. The first case is when $\angle ABC$ and $\angle ADC$ are supplementary angles (add up to 180 degrees), in which case $\angle BCD$ would be 0 degrees. The second case is when $\angle ABC$ and $\angle ADC$ are equal, in which case $\angle BCD$ would be 90 degrees.